For real numbers $x,$ what is the largest possible value of \[ 5 - (x-3)^2 ?\].

For solving this question, you can calculate the limit of f(x) for x→∞. Since the definition of limit is: the limit of f(x), as x approaches b , equals L. See: [tex]\lim _{x\to \ b \:}f(x))=L[/tex]

Therefore:

[tex]\lim _{x\to \infty \:}\left(5-\left(x-3\right)^2\right)\\ \\ \lim _{x\to \infty \:}\left(5\right)-\lim _{x\to \infty \:}\left(\left(x-3\right)^2\right)[/tex]

Applying the Constant Law - The limit of a constant is the constant itself: [tex]\lim _{x\to \infty \:}\left(5\right)=5[/tex]

Applying the Difference Law - The limit of a difference is the difference of the limits: [tex]\lim _{x\to \infty \:}\left(\left(x-3\right)^2\right)=\infty[/tex]

Thus,

[tex]\lim _{x\to \infty \:}\left(5-\left(x-3\right)^2\right)= \lim _{x\to \infty \:}\left(5\right)-\lim _{x\to \infty \:}\left(\left(x-3\right)^2\right)= 5- \infty=- \infty[/tex]

The largest possible value of f(x) will be -∞.

The result is consistent since it is possible to note that [tex](x-3)^2[/tex]positive number. Also, see that there is a difference between 5 and [tex](x-3)^2[/tex]-∞.